We want to count the cases that do not work (sum becomes less than 12).

Note that if 9 is one of the three digits then the sum of the digits will be 12 or more (least case is 9+1+2 =12).

Now if we restrict to the remaining five available digits, none of the cases work except {5, 4, 3} (as 5 + 4 + 3 = 12).

Therefore we select 3 distinct digits from the five digits and delete the one selection of {5, 4, 3}. This gives us the total number of bad cases.

$$ {{5}\choose {3}} – 1 = 9 $$

Hence the good selections are: total – bad = 20 – 9 =11.

Next note that each selection of three distinct integers can be permuted in 3!= 6 ways. Hence the total number of three -digit numbers that can be formed with the desired property = \( 11 \times 6 = 66 \)

Ans: (B) 66

2. If \( \sqrt{3} + 1 \) is a root of the equation \( 3x^3 + ax^2 + bx + 12 = 0 \) where a and b are rational numbers, then b is equal to

(A) -6; (B) 2; (C) 6; (D) 10;

Discussion:

Clearly if \( 1 + \sqrt {3} \) is one of the roots, then it’s conjugate \(1 – \sqrt {3} \) is another root (as all the coefficients are rational numbers).

Suppose the third root is \( \gamma \). (By Fundamental theorem of algebra there are three roots of a cubic).

By Vieta’s Theorem, the product of the roots is \( \frac{-12}{3} = -4\).

Hence:

$$ (1 + \sqrt{3} ) \times (1 – \sqrt {3}) \times \gamma = – 4 $$

This implies \( \gamma = 2 \) .

Now Vieta’s Theorem says, \( \frac{b}{3} \) is sum of the product of roots taken two at a time. Thus:

3. The sum of all integers from 1 to 1000 that are divisible by 2 or 5 but not divisible by 4 equals

(A) 175000; (B) 225500; (C) 149500; (D) 124000;

Discussion:

This is a problem related to Inclusion and Exclusion Principle in Combinatorics.

First lets add numbers from 1 to 1000 divisible by 2 but not by 4. So starting from 2 we add every fourth number. It is an arithmetic progression with first term 2, last term 998 and number of terms 250 (there are 500 even numbers from 1 to 1000 and \( \frac{1000}{4} = 250 \) is divisible by 4; hence the remaining 250 are divisible by 2 and not 4).

Therefore sum of the terms is:

$$ \frac {250}{2} (2 + 998) = 500 \times 250 = 125000 $$

Next we add the odd numbers that are divisible by 5 (we have already added the even ones).

So starting from 5, 15, 25, … , upto 995.

There are 100 such numbers and this is an arithmetic progression with common difference 10, first term 5 and last term 995.